Eng. & Tech. Journal, Vol.30, No.16, 2012
Absolute Maximum Bending Moment in Short Simply
Supported Beams under Moving Point Loads
Layla Ali Ghaleb Yaseen
Building and Construction Engineering Department, University of Technology/Baghdad
Email: laylayassin@yahoo.com
Received on: 4/9/2011 & Accepted on: 7/6/2012
ABSTRACT
This paper represents a theoretical investigation of the location of the absolute
maximum bending moment in short simply supported beams under the influence of
several moving point loads. However, for decades it was considered that the absolute
maximum bending moment is determined by positioning the beam center-line midway between the resultant of the loads and the nearer heavy load. Obviously, other
point loads may also be required to be mirror imaged with the resultant to ensure that
the obtained maximum bending moment is an absolute maximum. This method was
assumed to be working for all spans of simply supported beams without any
limitations.
In this paper it is shown that this traditional method is not always valid. Examples
of short span beams subjected to moving point loads having distance between two
loads exceeding half the length of the beam indicate absolute maximum bending
moments greater than those obtained by the traditional method.
Keywords: Absolute maximum bending moment; Short simply supported beams;
Moving point loads; location of maximum moment.
‫ﻋزم اﻹﻧﺣﻧﺎءاﻷﻗﺻﻰ اﻟﻣطﻠق ﻓﻲ اﻟﻌﺗﺑﺎت اﻟﺧرﺳﺎﻧﯾﺔ ذات‬
‫اﻹﺳﻧﺎد اﻟﺑﺳﯾط ﺗﺣت ﺗﺄﺛﯾر اﺣﻣﺎل ﻣرﻛزة ﻣﺗﺣرﻛﺔ‬
‫اﻟﺧﻼﺻﺔ‬
‫ھذا اﻟﺑﺣث ﯾﻘدم ﺗﺣﻠﯾﻼ ﻧظرﯾﺎ ﻟﻣوﻗﻊ ﻋزم اﻹﻧﺣﻧﺎء اﻷﻗﺻﻰ اﻟﻣطﻠق ﻓﻲ اﻟﻌﺗﺑﺎت اﻟﺧرﺳﺎﻧﯾﺔ اﻟﻘﺻﯾرة‬
‫ ﻣﻧذ ﻋﻘود ﻛﺎن اﻹﻋﺗﻘﺎد اﻟﺳﺎﺋد‬.‫ذات اﻹﺳﻧﺎد اﻟﺑﺳﯾط ﺗﺣت ﺗﺄﺛﯾر ﻣﺟﻣوﻋﺔ ﻣن اﻷﺣﻣﺎل اﻟﻣرﻛزة اﻟﻣﺗﺣرﻛﺔ‬
‫ان ﻋزم اﻻﻧﺣﻧﺎء اﻷﻗﺻﻰ اﻟﻣطﻠق ﯾﺗم اﺣﺗﺳﺎﺑﮫ ﻣن ﺧﻼل ﺟﻌل اﻟﺧط اﻟوﺳطﻲ ﻟﻠﻌﺗﺑﺔ ﻓﻲ وﺳ ط اﻟﻣﺳ ﺎﻓﺔ‬
‫ وﻗد ﯾﺗطﻠب اﻷﻣر ﺗﻛرار ﻣﺛل ھذه اﻟﻣﻧﺎظرة ﻟﻣﺣﺻﻠﺔ‬.‫ﺑﯾن ﻣﺣﺻﻠﺔ اﻟﻘوى ﻟﻼﺣﻣﺎل واﻗرب ﺣﻣل ﺛﻘﯾل ﻟﮭﺎ‬
.‫اﻷﺣﻣﺎل ﻣﻊ اﺣﻣﺎل اﺧرى ﻟﺿﻣﺎن اﻟﺣﺻول ﻋﻠﻰ اﻟﻌزم اﻷﻗﺻﻰ اﻟﻣطﻠق‬
‫ھذه اﻟطرﯾﻘﺔ ﻛﺎن ﻣن اﻟﻣﻔﺗرض ان ﺗﻌﻣل ﻟﻛل اﻟﻌﺗﺑﺎت ذات اﻹﺳ ﻧﺎد اﻟﺑﺳ ﯾط وﺑﻣﺧﺗﻠ ف اﻷط وال وﺑ دون‬
.‫ﻣﺣددات‬
‫ اﻣﺛﻠ ﺔ ﻣ ن ﻋﺗﺑ ﺎت ذات اط وال‬.‫ﻓﻲ ھ ذا اﻟﺑﺣ ث ﺗ م ﺗوﺿ ﯾﺢ ان اﻟطرﯾﻘ ﺔ اﻟﺗﻘﻠﯾدﯾ ﺔ ﻗ د ﻻ ﺗﺻ ﺢ داﺋﻣ ﺎ‬
‫ﻗﺻﯾرة ﻣﻌرﺿﺔ ﻻﺣﻣﺎل ﻣﺗﺣرﻛﺔ ﻓﯾﮭﺎ اﻟﻣﺳﺎﻓﺔ ﺑﯾن ﺣﻣﻠ ﯾن ﻣﻧﮭ ﺎ ﯾزﯾ د ﻋﻠ ﻰ ﻧﺻ ف ط ول اﻟﻌﺗﺑ ﺔ ﺗﻌط ﻲ‬
.‫ﻋزم اﻧﺣﻧﺎء اﻗﺻﻰ ﻣطﻠق اﻋﻠﻰ ﻣن اﻟﻌزم اﻟﻣﺣﺗﺳب ﺑﺎﻟطرﯾﻘﺔ اﻟﺗﻘﻠﯾدﯾﺔ‬
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
NOTATIONS
Mabs : Absolute Maximum Bending Moment
MCL
: Moment at Mid-span of the Beam
R
: P1/P2, (r>1)
P1
: Heaviest Moving Point Load
P2
: Load adjacent to Heaviest Moving Point Load
Pl1,
Pl2,
: Other Moving Point Loads
Pli
x
: Distance between Heaviest Load and Resultant
R
: Resultant of moving point loads
L
: Length of Beam
d,
dl1,
dl2,
: Distances between Point Loads and Heaviest Load
dli
dcr
: Critical Distance between moving point loads
dout
: Minimum Distance that keeps P 2 within the vicinity of the beam
w
: d/L
wcr
: dcr /L
wout
: dout /L
S
: R/P2
S1
: ∑P li dli /P2 L
S2
: ∑P li /P2
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Eng. & Tech. Journal, Vol.30, No.16, 2012
Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
INTRODUCTION
or simply supported beams the critical positions of the loading and the
associated absolute maximum bending moment cannot, in general, be
determined by inspection.
Russell C. Hibbeler [1] concluded that the critical position of the absolute
maximum moment in a simply supported beam occurs under one of the
concentrated forces such that this force is positioned on the beam so that it and the
resultant force of the system are equidistant from the beam's center line. As a general
rule, the absolute maximum moment often occurs under the highest force lying
nearest to the resultant force of the system.
W. R. Spillers [2] developed a simple result which can be useful in determining
how loads should be positioned in order to produce maximum bending moment. He
selected one load of a group of moving loads and found the location of the load group
so that the moment under that load is considered to be the maximum. For maximum
effect, the load group should be positioned so that the distance from the centroid of
the load group to the right support equals the distance from the left support to the load
under consideration.
R. L. Jindal [3] considered that the moment at the center is generally not the
absolute maximum bending moment and in short beams the difference is appreciable.
The maximum moment under any load will occur when that load and the center of
gravity of all the loads are equidistant from the center of the beam.
Research Significance
The present theoretical investigation examines some limitations in the calculation
and location of the absolute maximum bending moment in short simply supported
beams; since most of the moving loads in Iraq are due to vehicles moving on roads,
and the main vehicle characteristics are limited by the state or highway authorities to
suit both transportation services and highway network conditions. It states that the
maximum distance between wheels in a semitrailer combination is limited to (7.93
m)[4], therefore, the term short beam in this research will refer to beams with spans
less than (15 m).
F
THEORETICAL INVESTIGATION
Two types of loadings are considered in this research, two point loads and more
than two point loads.
Simply Supported Beams Under The Influence of Two Moving Point Loads
A simply supported beam of length (L) loaded with two moving point loads (P 1)
and (P2 ) where (P1) is the heaviest load; by applying the traditional absolute
maximum bending moment procedure, the resultant force would be ( R = P1 + P2 )
and the location of the absolute maximum moment would be under one of the point
loads such that this force is positioned on the beam so that it and the resultant force of
the system are equidistant from the beam's mid-span, as shown in Figure (1).
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
x
CL R
L/2-x/2
P1 x/2
R
P1
P2
P2
A
d
B
L
P1
Mabs
A
V1
RA
L/2-x/2
Figure (1) – A Simply Supported Beam Loaded with
Two Point Loads with (d>L/2).
R = P1 + P2 and M1 = P2 d
x=
RA =
M1 P2d
=
R
R
R  L x  R  L− x 
 − = 

L 2 2  L 2 
M abs =
R
( L − x )2
4L
If (d> L/2) then the moment at the center of the beam due to (P1) would be:
MCL= P 1L/4.
The relationship between Mabs and MCL would be:
R
2
(L x )2 M
M abs 4 L −
r+1
w 
=
⇒ abs =
1 −

P1 L
M CL
M CL
r 
r+1
4
Where: r = P1 , w = d and r ≥ 1 ,
P2
L
The relationship between (Ma bs/MCL) and (w) is shown in (Fig. 2); the error area in
the traditional absolute maximum bending moment procedure lies under the line
(Mabs
MCL = 1)
.
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
Mabs / M CL
3
1r =
1.5r =
2r =
2.5r =
3r =
1Mabs/MCL=
2
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
w=d/L
Figure (2) – The Relationship between (Mabs/MCL) and (w).
The critical distance (dcr ) between the two moving point loads can be obtained by
substituting
Mabs = MCL
R
( L − x )2 = P 1 L
4L
4
Pd
Then by substituting, ( R = P1 + P2 ) and ( x = 2 ) the following is obtained:
R
L( P1 + P2 ) P2 × d
P22 × d 2
P ×L
−
+
= 1
4
2
4 L( P1 + P2 )
4
 ( P + P2 ) ± ( P1 + P2 )× P1
d = 1

P2


× L


Since the minimum d is required, and R is much greater than P 2, then the positive
sign will be ignored and dcr would be:
 (P + P2 ) − ( P1 + P2 )× P1
dcr =  1

P2

(
dcr = r + 1 r + 1 − r

× L


)
……. (1)
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
wcr =
d cr
= r+1×
L
(
r+1− r
)
…… (2)
Any distance between the two point loads larger than the amount calculated by Eq.
(1), (dcr ), leads to a failure in the traditional absolute maximum moment procedure
and the maximum moment location would be at the middle of the beam under the
influence of the heaviest point loads (P 1).
If the distance ( d >
L x
+ ), then P2 will be positioned outside the beam, therefore
2 2
the minimum distance (dout ) that keeps (P2) within the vicinity of the beam would be:
d out =
wout =
r+1
×L
2r + 1
dout r + 1
=
L 2r + 1
…… (3)
d
< wout
L
wcr <
r+1×
(
)
r+1− r <
d r+1
<
L 2r + 1
The distance between the moving point loads with relative to the length of the
beam depends on the value of (r) which is always greater than or equal to (1).
(
)
wc = f (r) = r + 1 × r + 1 − r , r ≥ 1
If (r = 1) then:
f (r ) = 0.5858
If (r →∞ ) then:
Lim (r → ∞ ) f (r ) = 0 .5
This relationship is shown in (Figure 3).
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
0.60
@r=1;
wcr= 0.5858
f ( r ) = wcr
0.58
0.56
0.54
0.52
0.50
0
5
10 15 20 25 30 35 40 45 50 55
(r=P1 /P2)
Figure (3) – The Relationship between (r=P1/P2) and (f (r) = wcr)
According to the above function, in the case of two moving point loads if the
distance between the two point loads (d) accedes 0.5L, no matter what the values of
(P1) and (P 2) are; if (wcr < w< wout ), then the traditional absolute maximum moment
procedure is invalid and the maximum moment occurs when the heaviest point load is
positioned at mid-span.
Figure (4) shows the validity of the traditional absolute maximum moment
procedure for a simply supported beam with regard to w, wcr and wout .
w =d/L
0.8
wout
wcr
wcr < w < wout
0.7
0.6
0.5
0.4
0
1
2
3
4
5
6
7
8
9
10
r =P1/P2
Figure (4) – The Validity of the Traditional Absolute Maximum
Moment Procedure for a Simply Supported Beam.
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
SIMPLY SUPPORTED BEAMS UNDER THE INFLUENCE OF MORE THAN
TWO MOVING POINT LOADS
For a simply supported beam of length (L) under the influence of more than two
moving point loads (P 1, P2, P l1, P l2,….,Pli ), three cases must be considered regarding
the location of the heaviest moving point load, these cases are:
a)
If the heaviest load is in the middle, then the traditional absolute
maximum moment procedure may be applied to calculate the maximum
bending moment.
b)
If the heaviest load is the second load, then the resultant would be
nearer the heaviest load; by applying the traditional absolute maximum
moment procedure, the resultant force would be ( R = P1 + P2 + Pli ) and the
∑
location of the absolute maximum moment would be under the heaviest closer
point load ( P1 ) such that this force is positioned on the beam so that it and the
resultant force of the system are equidistant from the beam's mid-span, as
shown in (Figure 5).
Pl2 Pl1 P1 x
R
CL
x/2 R
P2
dl1
A
d
dl2
Pl2 Pl1 P1
dl1
dl2
L/2
L/
A
MCL =
B
P1 L
+
4
∑P
li
2
L

 − dli 
2


B
d
L/2
Pl2 Pl1 P1
dl1
P2
Pl2 Pl1 P1
P2
A
dl2 V
1
R L/2-x/2
Mab
Figure (5) - A Simply Supported Beam Loaded with more
than Two Point Loads, the 2nd Load is the Heaviest.
R = P1 + P2 + ∑ Pli
M = P2d - ∑Plidli
x=
M P2 d - ∑ Pli dli
=
R
R
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
Mabs =
R
(L − x )2 − ∑ Pli dli
4L
But:
M CL =
P1 L
+
4
∑P
li
2
L

×  − d li 
2


Then by substituting Mabs =MCL
(
2
)
d
d
2
2
  − 2(S + S1 )   + S + S1 − Sr − SS2 = 0
 L
 L
d
  = wcr = ( S + S 1 ) − S (2 S 1 + S 2 + r )
 L
……(4)
Where:
S=
R
, S1 =
P2
∑P d
li
P2 L
li
, S2 =
∑P
li
P2
and r =
P1
P2
To calculate wout=dout /L, substitute d=(L/2+x/2)
dout =
∴
L x
+
2 2
S - S1
d out
= wout =
2S −1
L
…… (5)
So if (wcr < w = d/L < wout) , then the traditional absolute maximum moment
procedure does not work and the maximum moment occurs when the heaviest load is
positioned at the center of the beam.
c) If the heaviest load is the first load, then two cases must be tested:
− When x< d/2:
In this case the resultant of the system would be closer to the heaviest load.
By applying the traditional absolute maximum moment procedure, the
resultant force would be:
( R= P1 + P2 + Pli ) and the location of the absolute maximum moment would be
∑
under the heaviest point load ( P1 ) such that this force is positioned on the beam so
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
that it and the resultant force of the system are equidistant from the beam's midspan, as shown in Figure (6).
x
P1
CL R
x/2
R
P2
d
dl1
Pl1 Pl2
P2
P1
A
L/2
dl2
CL
P1
d
P2
A
B
L/2
L/2
B
L/2
d
Pl1 Pl2
dl1
dl2
P1
A
RA
L/2-x/2
V1
Mabs
Figure (6) - A Simply Supported Beam Loaded with more than Two
Point Loads the 1st Load is Heaviest - Case1 (x<d/2).
R= P1 + P2 + ∑ Pli
M = P2 d + ∑ Plidli
x=
Mabs =
M P2 d + ∑ Pli dli
=
R
R
R
( L − x )2 , MCL = P1 L
4L
4
Then by substituting Mabs=MCL
R
( L − x)2 = P1 L
4L
4
 d  2
d 
  − 2 ( S − S1 )  + 
 L 
 L 


2
( S − S1 ) − Sr = 0

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Eng. & Tech. Journal, Vol.30, No.16, 2012
Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
d
= w cr = ( S − S 1 ) −
L
Where:
S=
Sr
……(6)
R
∑ Plidli and r = P1
, S1 =
P2
P2 L
P2
To calculate wout=dout/L, substitute d=(L/2+x/2)
L x
dout = +
2 2
∴
dout
S + S1
= wout =
L
2S − 1
…… (7)
So if (wcr < d/L < wout ) , then the traditional absolute maximum moment procedure
does not work and the maximum moment occurs when the heaviest load is positioned
at the center of the beam.
− When x> d/2:
In this case the resultant of the system would be closer to the second load
(not the heaviest load), and that happen when the value of the second load is
approaching the value of the heaviest load; by applying the traditional
absolute maximum moment procedure, the resultant force would be:
( R = P1 + P2 + Pli ) and the location of the absolute maximum moment
∑
would be either under the closer point load ( P2 ) or under the farther point
load ( P1 ) whichever the heaviest as shown below:
a) The closer point load ( P2 ) is positioned on the beam such that this force and
the resultant force of the system are equidistant from the beam's mid-span,
as shown in (Fig.7).
R = P1 + P2 + ∑ Pli
M( P1 ) = P2d + ∑ Plidli
x=
M P2 d + ∑ Pli dli
=
R
R
RB =
R L d − x 
 −

L 2
2 
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Eng. & Tech. Journal, Vol.30, No.16, 2012
P1
R
x
d
P2
Pl1 Pl2
dl1
dl2
Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
R CL (d-x)/2
P2 Pl1 Pl2
x
P1
A
d
L/4
P1
A
dl1
dl2
P2 Pl1 Pl2
P2 Pl1 Pl2
d
B
L/2
L/2
B
dl1
dl2
Mabs
B
V1
L/2-[(d-x)/2] RB
L
Figure (7) - A Simply Supported Beam Loaded with more than Two
Point Loads the 1st Load is Heaviest -Case2-a (x>d/2).
M abs =
R
(L − d + x )2 − ∑ Pli (dli − d )
4L
M CL =
P2 L
+
4
∑P
……(8)
L

 − (d li − d )
2

li
2
Then by substituting Mabs=MCL
R
( L − d + x )2 − ∑ Pli (dli − d ) = P2 L + ∑ Pli  L − (dli − d ) × 1
4L
4
2
 2
(S − 1)(S + S1 ) d 
2
2
−2 
  + S + S1 − S (1 + S2 ) = 0
−
L
SS
L
 
2

 
S (2 S 1 + S 2 + 1 )
−
(S − 1 )2
(S + S 1 ) − SS 2 − 2 SS 2 (S + S 1 ) +
=
(S − 1 ) (S − 1 )2
… (9)
(S − 1 )3
(S − 1)2  d 
d
= w cr
L
(
2
)
2
S 2 S2
(S − 1 )4
Where:
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
S=
Pd
P
R
P
, S1 = ∑ li li , S 2 = ∑ li and r = 1
P2
P2 L
P2
P2
To calculate wout =dout /L, substitute d=[L/2+(d-x)/2]
d out =
L+d − x
2
d out
(S − S 1 )
= wout =
(S + 1 )
L
……(10)
So if (wcr < d/L < wout ) , then the traditional absolute maximum moment procedure
does not work and the maximum moment occurs when (P2) is positioned at the center
of the beam.
b) The farther point load ( P1 ) is positioned on the beam such that this force and the
resultant force of the system are equidistant from the beam's mid-span, as shown
in (Figure 8).
P1
CL R
R
x
P2
d
Pl1 Pl2
A
dl1
d
L/4
A
dl1
dl2
P1
P2 Pl1 Pl2
d
B
L/2
L/2
dl2
P1
P2 Pl1 Pl2
P1 x/2 x/2
B
dl1
dl2
A
RA
L/2-x/2d
V1
Mabs
L
Figure (8) - A Simply Supported Beam Loaded with more than Two Point
Loads the 1st Load is Heaviest-Case2-b (x>d/2).
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
R = P1 + P2 + ∑ Pli
M = P2 d + ∑ Plidli
x=
M P2 d + ∑ Pli dli
=
R
R
RA =
Mabs =
R L x 
 − 
L 2 2
R
(L − x ) 2
4L
M CL =
……(11)
P2 L
L
 1
+ ∑ Pli  − (dli − d ) ×
4
2
 2
Then by substituting Mabs=MCL
P
R
( L − x )2 = P2 L + ∑ li
4L
4
2
[(
2
L

 − (dli − d )
2

)
]
d
d 
2
2
  − 2[S (S2 + 1) − S1 ]  + S + S1 − S (S2 + 1) = 0
 L
 L
S 2 S2 ( S2 + 2 )
d
= wcr = S (S2 + 1) − S1 −
L
− S(S2 + 1)(2 S1 − 1)
…(12)
Where:
S=
Pd
P
R
, S1 = ∑ li li and S 2 = ∑ li
P2
P2 L
P2
To calculate wout=dout/L, substitute d=[L/2+x/2]
dout S + S1
=
L 2S − 1
… (13)
So if (wcr < d/L < wout ) , then the traditional absolute maximum moment procedure
does not work and the maximum moment occurs when the second point load is
positioned at the center of the beam.
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Eng. & Tech. Journal, Vol.30, No.16, 2012
Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
CONCLUSIONS AND RECOMMENDATIONS
From the theoretical investigation done in this research the following can be
concluded:
1) For short simply supported beams subjected to two point loads, if the distance
between these two loads is more than half the length of the beam, then the
maximum bending moment occurs when the heaviest load is positioned at midspan.
2) For such beams (wcr) and (wout ) must be calculated according to Eq. (2) and Eq.
(3) respectively. If (wcr<w<wout ), then the traditional absolute maximum
moment procedure is invalid and the absolute maximum moment occurs when
the heaviest load is positioned at mid-span.
3) For short simply supported beams subjected to more than two point loads, and
the distance between the heaviest load and the load adjacent to is more than
half the length of the beam, then three cases must be checked:
a) If the heaviest load occurs in the middle of the moving loads, then the
traditional absolute maximum bending moment procedure may be applied
to calculate the amount and location of the absolute maximum bending
moment.
b) If the heaviest load is the second load from either side of the beam, then
(wcr) and (wout ) must be calculated according to Eq.(4) and Eq.(5)
respectively, to check the validity of the traditional absolute maximum
bending moment procedure.
If (wcr<w<wout), then the traditional absolute maximum bending moment
procedure would be invalid and the absolute maximum moment occurs
when the heaviest load is positioned at mid-span.
c) If the heaviest load is the first load from either side of the beam, then two
cases must be checked:
−
If the distance between the first load and the resultant of system of moving
loads (x) is less than half the distance between the first two loads (x<d/2),
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Eng. & Tech. Journal, Vol.30, No.16, 2012
Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
then (wcr) and (wout) must be calculated according to Eq.(6) and Eq.(7)
respectively, to check the validity of the traditional absolute maximum
bending moment procedure. If (wcr<w<wout), then the absolute max moment
theory would be invalid and the maximum moment occurs when the
heaviest load is positioned at mid-span.
− If the distance between the first load and the resultant of moving loads (x) is
more than half the distance between the first two loads (x>d/2), the resultant
would be close to the second load (not the heaviest one), then two cases
must be checked for the absolute maximum moment location:
∗ Case-1: When the resultant is positioned on the beam such that it and the
closest load (not the heaviest one) are equidistant from the beam's mid-span,
in this case Mabs must be calculated according to Eq. (8).
∗ Case-2: When the resultant is positioned on the beam such that it and the
farther load (the heaviest one) are equidistant from the beam's mid-span,
Mabs must be calculated according to Eq. (11).
If the moment in case-1 is larger than the moment in case-2, then case-1 would
be dominating and (wcr) and (wout ) must be calculated according to Eq. (9) and
Eq. (10) respectively, to check the validity of the traditional absolute maximum
bending moment procedure. If (wcr<w<wout ), then the traditional absolute
maximum bending moment procedure would be invalid and the maximum
moment occurs when the second point load (P2) is positioned at mid-span.
Otherwise case-2 would be the dominating case, then (wcr) and (wout ) must be
calculated according to Eq. (12) and Eq. (13) respectively, to check the validity
of the traditional absolute maximum bending moment procedure. If
(wcr<w<wout ), then the traditional absolute maximum bending moment
procedure would be invalid and the maximum moment occurs when the second
point load (P2) is positioned at mid-span.
4) In short simply supported beams, designers must check the correct location and
amount for the absolute maximum bending moment, following the cases
mentioned in this research; the highest moment must be taken into
consideration for the design.
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
APPLICATIONS
1) Example on short simply supported beam subjected to two point loads:
− Calculate the location and amount of the absolute maximum bending moment in
the simply supported beam of 5m length subjected to the moving loads shown in
the figure,belov:
kN
kN
30
20
3m
Upon applying the method proposed in this research according to Eq.(2) and
Eq.(3), the following is found:
r=
P1 30
=
= 1.5
P2 20
w=
d 3
= = 0.6
L 5
(
)
wcr = r + 1 × r + 1 − r = 0.625
wout =
r+1
= 0.625
2r + 1
Since ( wcr < w < wout ), then the traditional absolute maximum moment
procedure is invalid, therefore the maximum bending moment occurs when the
heaviest load is positioned at mid-span:
Mmax =
PL (30)(5 )
=
= 37.5 kN .m
4
4
To check that, below is the calculation of moment according to the traditional
absolute maximum bending moment procedure:
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Eng. & Tech. Journal, Vol.30, No.16, 2012
Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
1.2m 50 kN
P1
kN
kN
30
20
3m
CL
m R
0.6
P2
A
B
5
m
L/2-x/2
kN
30
Mab
A
V
RA
m
2.5-0.6=1.9
R = 30 + 20 = 50 kN
M = 20 × 3 = 60 kN .m
x =
60
x
= 1 .2 m ,
= 0 .6 m
50
2
2
M abs =
2)
R L x
×  −  = 36 .1 kN .m
L 2 2
Examples on short simply supported beam subjected to more than two moving
point loads:
a) Calculate the location and amount of the maximum bending moment in the
simply supported beam of 6m length subjected to the moving loads shown in
figure.
0.5m R
10kN
10kN
1m
20kN
30kN
1m
3.2m
The second load from the right of the moving loads given is the heaviest one; upon
applying the method proposed in this research using Eq. (4) and Eq. (5) the following
is found:
R = 10 + 10 + 30 + 20 = 70 kN
M(P =30kN ) = 34kN .m
1
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Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
34
x
= 0.5m, = 0.25m
70
2
d 3 .2
w=
=
= 0 . 53 ,
L
6
70
30
S=
= 3. 5 , r =
= 1.5
20
20
(10 ) (2 ) + (10 ) (1)
(10 + 10 ) = 1
S1 =
= 0.25 , S 2 =
(20 ) (6 )
20
x=
wcr = ( S + S 1 ) − S (2 S 1 + S 2 + r ) = 0.51
S - S1
wout =
= 0.54
2S − 1
wcr < w < wout
Then the traditional absolute maximum moment procedure is invalid, and the max
moment occurs when (P1) is positioned at mid-span.
M max =
P1 L ∑ Pli  L

+
 − dli 
4
2 2

M max = 45 + 15 + 5 = 65 kN .m
While the moment according to the traditional absolute maximum procedure will
be:
x-2/2 CL
10
kN
10
kN
kN
30
R
k
20
A
B
3m
10
kN
10
3m
kN
kN
30
A
RA
Mabs
V
3-0.25=2.75m 1
RA =
M abs
70 (3 − 0.25 )
= 32.1 kN .m
6
= 58.275 kN .m
b) Calculate the location and amount of the maximum bending moment in the simply
supported beam of 5m length subjected to the moving loads shown in figure.
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Eng. & Tech. Journal, Vol.30, No.16, 2012
Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
c)
10kN
30kN
10kN
1m
3m
The first load of the moving loads given is the highest one; upon applying the
method proposed in this research using Eq. (6) and Eq. (7) the following is found:
R = 30 + 10 + 10 = 50 kN
M( P1 =30kN ) = 70 kN.m
70
d
= 1.4 m , = 1.5 m ,
2
50
d
∴x <
2
x=
w=
S=
50
= 5,
10
S1 =
d 3
= = 0 .6
L 5
(10 ) (4 )
= 0.8 ,
(10 ) (5 )
r=
30
=3
10
wcr = (S − S1 ) − Sr = 0.33
S + S1
wout =
= 0.64
2S − 1
wcr < w < wout
Then the traditional absolute maximum moment procedure is invalid, and the max
moment occurs when P1 is positioned at mid-span.
Mmax = MCL =
P1 L
= 37.5 kN .m
4
d) Calculate the location and amount of the maximum bending moment in the simply
supported beam of 6m length subjected to the moving loads shown in figure.
30 kN
R
x=2.4m
3.2m
20kN
15kN
1m
10kN
1m
The first load of the moving loads given is the highest one; first of all check x and
d/2:
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Absolute Maximum Bending Moment in
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Moving Point Loads
Eng. & Tech. Journal, Vol.30, No.16, 2012
R = 30 + 20 + 15 + 10 = 75 kN
M( P1 =30kN ) = 179 kN .m
d 3.2
179
= 2.4 m ,
=
= 1.6 m ,
2
2
75
d
∴x >
2
x=
Upon applying the method proposed in this research, two cases must be checked:
− Case-1: When the resultant is positioned on the beam such that it and the closest
load (P 2) are equidistant from the beam's mid-span; calculate Mabs according to
Eq. (8):
M abs =
R
(L − d + x )2 − ∑ Pli (d li − d ) = 49 .5 kN .m
4L
− Case-2: When the resultant is positioned on the beam such that it and the heaviest
load (P1) are equidistant from the beam's mid-span, calculate Mabs according to Eq.
(11):
Mabs =
R
( L− x )2 = 40.5 kN .m
4L
Mabs (case − 1 ) ⟩ Mabs (case − 2 )
Then case-1 is dominating, therefore using Eq. (10) and Eq. (11) the following is
found:
d 3.2
w= =
= 0.53
L 6
S=
(15)(4.2) + (10)(5.2) = 0.96, S = (15 + 10) = 1.25, r = 30 = 1.5
75
= 3.75, S1 =
2
(20)(6)
20
20
20
wcr =
(S + S 1 ) − SS 2
(S − 1) (S − 1)2
S (2 S 1 + S 2 + 1 )
(S − 1)2
2 SS 2 (S + S 1 )
= 0.52 ,
− −
(S − 1)3
2
+
S 2 S2
(S − 1)4
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wout =
(S − S 1 ) = 0.59
(S + 1 )
Eng. & Tech. Journal, Vol.30, No.16, 2012
Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
wcr < w < wout
Then the traditional absolute maximum bending moment procedure is invalid, and
the max moment occurs when (P2 ) is at mid-span.
Mmax =
P2 L
P L

+ ∑ li  − (dli − d)
4
2 2

Mmax = 50 kN .m
A Summary of the Applications’ Results and the Percentage of Error between the
Traditional Method and the Proposed Method is shown in Table-1, while a
comparison between the traditional method and the proposed method is shown in
Table-2.
Table (1) – Shows the Percentage of Error between the
Traditional Method and the Proposed Method
MT
MP
% of
Cases
kN.m kN.m Error
Simply Supported Beam
36.1
37.5
3.7
(2-Point Loads)
Simply Supported Beam
(3-Point Loads)
26.8
37.5
28.5
1st Load is the Heaviest (x<d/2)
Simply Supported Beam
(4-Point Loads)
58.2
65
10.5
2st Load is the Heaviest
Simply Supported Beam
(4-Point Loads)
49.5
50
1
1st Load is the Heaviest (x>d/2)
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Eng. & Tech. Journal, Vol.30, No.16, 2012
Absolute Maximum Bending Moment in
Short Simply Supported Beams under
Moving Point Loads
Table(2) – Shows a Comparison between the Traditional
Method and the Proposed Method
Cases
Method Used
d ≤ L/ 2
Simply
Supported
Beam
(Two Loads)
d > L/ 2
Proposed Method:
Traditional Method
Calculate wcr and wout using Eqs. (2&3)
wcr < w < wout
Mabs = When P1 is located at Mid-Span
Simply
Supported
Beam
( > Two Loads)
Heaviest Load in the
Middle
Simply
Supported
Beam
( > Two Loads)
2nd Load is the
Heaviest
Traditional Method
d ≤ L/ 2
d > L/ 2
Proposed Method:
Traditional Method
Calculate wcr and wout using Eqs. (4&5)
wcr < w < wout
Mabs = When P1 is located at Mid-Span
d ≤ L/ 2
Traditional Method
Proposed Method:
L
d
d>
and x <
2
2
Simply
Supported
Beam
( > Two Loads)
1st Load is the
Heaviest
Calculate wcr and wout using Eqs. (6&7)
wcr < w < wout
Mabs = When P1 is located at Mid-Span
d>
If Mabs (Eq.8) is Dominating calculate wcr
and wout using Eqs. (9&10)
L
d
and x >
2
2
wcr < w < wout
Calculate Mabs using
Mabs = When P 2 is located at Mid-Span
Eqs. (8&11)
If Mabs (Eq.11) is Dominating calculate wcr
and wout using Eqs. (12&13)
The largest moment
wcr < w < wout
will dominate
Mabs = When P 2 is located at Mid-Span
REFERENCES
[1].Hibbler, R. C. "Structural Analysis", 2nd ed. Macmillan, New York, (1990).
[2].Spillers, W. R. "Introduction to Structures". Ellis Horwood, West Sussex,
England, (1985).
[3].Jindal, R. L. "Elementary Theory of Structures", Third Edition, S. Chand &
Company LTD, Ram Nagar, New Delhi-110055, (1986).
[4].Republic of Iraq, Ministry of Housing and Construction, State Organization of
Roads and Bridges, Highway Design Manual, Design & Studies Dept. Road &
Traffic Division, 1982.
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